Two $\mathfrak{T}(k)$-Summation Equivalences of the Riemann Hypothesis (The Riemann Hypothesis and Young´s Lattice [Part 4/9])

Two T(k)-Summation Equivalences of the Riemann Hypothesis José Damián Espinosa December 13, 2025 Dedication: Στην αγαπημένη μου Θεά Μαρσέλα (A mi amada Diosa Marcela) Abstract In this article, Two T(k)-Summation Equivalences of the Riemann Hypothesis are proposed, making direct use of the EspinosaRiemann Equivalence Theorem. These equivalences take the function T(k)as a summation respectively, involving generalized harmonic numbers in sequences, yielding two beautiful versions from this T(k)-Summation approach. The Riemann Hypothesis is equivalent to: H1 n 1! 1 1H(1) 1+H2 n 2! 1 1H(1) 2−1 2H(2) 1+· · · > σ(n) for all n∈N, where H(m) p=Pp j=1 1 jmdenotes the p-th generalized harmonic number of order m,Hn=Pn i=1 1 iis the n-th harmonic number and σ(n) = Pd|ndis the sum-of-divisors function of n. The Riemann Hypothesis is equivalent to: H1 n 1! 1 1h(1) 2+H2 n 2! 1 1h(1) 3+1 2h(2) 3+· · · > σ(n) for all n∈N, where h(p) m=Pm j=2 1 jpdenotes the m-th generalized harmonic number of order pstarting from j= 2,Hn=Pn i=1 1 iis the n-th harmonic number and σ(n) = Pd|ndis the sum-of-divisors function of n. 1 “The essence of mathematics is not to make simple things complicated, but to make complicated things simple.”– S. Gudder “Everything should be made as simple as possible, but no simpler.”– Albert Einstein “Truth is ever to be found in simplicity, and not in the multiplicity and confusion of things.”– Isaac Newton “Simplicity is the ultimate sophistication.”– Leonardo da Vinci “Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of a sculpture, without any appeal to our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.”– Bertrand Russell “Beauty is the first test: there is no permanent place in the world for ugly mathematics.”– G. H. Hardy “A mathematician is not complete until he is a little bit of a poet in his soul.”– Sofia Kovalevskaya “The scientist does not study nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.”– Henri Poincaré “Mathematics, rightly viewed, possesses not only truth, but supreme beauty.”– Bertrand Russell 2 “Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.”– Albert Einstein “Logic will get you from A to B. Imagination will take you everywhere.”– Albert Einstein “If I had an hour to solve a problem, I’d spend 55 minutes thinking about the problem and 5 minutes thinking about solutions.”– Albert Einstein “An expert is a person who has made all the mistakes that can be made in a very narrow field.”– Niels Bohr “To attain the impossible, one must attempt the absurd.”– Miguel de Cervantes “We must know, we will know ( Wir müssen wissen. Wir werden wissen.).”– David Hilbert Contents 1 Introduction 4 2 Content 4 2.1 Two T(k)-Summation Equivalences of the Riemann Hypothesis . 4 2.1.1 Preliminary Concepts . . . . . . . . . . . . . . . . . . . . 5 2.1.2 On Pk∈N(Hk−γ)xk k!..................... 6 2.1.3 On the Espinosa-Riemann Equivalence Theorem . . . . . 7 2.1.4 On the Riemann zeta function ζand the Euler-Mascheroni constant γ........................... 8 2.1.5 On some useful inequalities . . . . . . . . . . . . . . . . . 9 2.1.6 On η(k)and its upper and lower bounds . . . . . . . . . . 12 2.1.7 On Ξ(x):=Pk∈N xk k!η(k).................. 20 3 2.1.8 Espinosa’s Riemann Hypothesis Equivalence (2025) with T(k)=η(k).......................... 26 2.1.9 On another important relationship between the Riemann zeta function ζand the Euler-Mascheroni constant γ. . . 33 2.1.10 On other useful inequalities . . . . . . . . . . . . . . . . . 34 2.1.11 On τ(k)and its upper and lower bounds . . . . . . . . . . 39 2.1.12 On κ(x):=Pk∈N xk k!τ(k).................. 45 2.1.13 Espinosa’s Riemann Hypothesis Equivalence (2025) with T(k)=τ(k).......................... 51 References 57 1 Introduction In this work, two equivalences of the Riemann Hypothesis are presented, resulting from the application of the Espinosa-Riemann Equivalence Theorem. This theorem, the main result of the article On a Theorem of Equivalences of the Riemann Hypothesis [Esp25b], proposes a general classical form for a set of equivalences of the Riemann Hypothesis, a certain type, shown below: O(Hn) = X k∈N Hk n k!T(k)> σ(n) This inequality applies for all n∈N, where Hn=Pn i=1 1 iis the n-th harmonic number, σ(n) = Pd|ndis the sum-of-divisors function of n, and the functions O:R+→Rand T:R+→Rmust satisfy three specific conditions that are discussed in detail and stated in Section 2.1.3. We make use of this result and propose two expressions very similar to each other. Although they derive and essentially share the same central idea as the main version proposed in the article A Beautiful Equivalence of the Riemann Hypothesis [Esp25a] (which used T(k)=M(k) = Pk2 i=k1 i). The novelty lies in defining T(k)cleanly by two expressions of summations of respectively weighted generalized harmonic numbers. This specific structure is referred to as the T(k)-Summation approach. This investigation is made possible thanks to the generalized framework proposed in the Espinosa-Riemann Equivalence Theorem. 2 Content 2.1 Two T(k)-Summation Equivalences of the Riemann Hypothesis In this Section 2.1, we work with a continuous interweaving of results until reaching Theorem 2 and 5, the two new proposed versions of the Riemann 4 Hypothesis respectively. A set of ordered subsections are presented that provide foundation as the narrative progresses, in a constructive manner, following the clear natural line that finally leads to the two long-awaited results mentioned. All the necessary results for the first equivalence are addressed, until reaching it, and immediately after, the next one begins, developing all the missing elements, in a completely analogous way to the first one, until leading to it, the second one. 2.1.1 Preliminary Concepts The present Section 2.1.1 addresses key concepts, which provide foundation for everything undertaken, for that which is constructed later. We begin by defining the famous Euler-Mascheroni Constant γ, an essential constant when addressing the Riemann Hypothesis, the definition follows: Definition 1 (Euler-Mascheroni Constant γ).The Euler-Mascheroni constant γis defined by the limit (Abramowitz [AS64, Formula 6.1.3]): γ:= lim n→∞   n X k=1 1 k−log n = 0.5772156649 ... This limit exists and is finite, representing the asymptotic difference between the harmonic series and the natural logarithm. Proposition 1 (Partial integral of the integral form of the Euler-Mascheroni constant γ).For all k∈N, the following holds: Z∞ k1 ⌊t⌋−1 tdt =γ−Hk−1+ log k where γis the Euler-Mascheroni constant and Hk−1=Pk−1 n=1 1 n(H0:= 0). Proof. See in Espinosa [Esp25c] (Two T(k)-Integral Equivalences of the Riemann Hypothesis). Proposition 2. For all k∈N: 1 k(k+ 1) <Z∞ k1 ⌊t⌋−1 tdt < 1 k Proof. See in Espinosa [Esp25c] (Two T(k)-Integral Equivalences of the Riemann Hypothesis). Having these results, we proceed to mention and define some fundamental concepts, such as those of convergence and radius of convergence of real series as follows in Definition 2: Definition 2 (Convergence of real power series).Let P∞ k=0 ckxkbe a power series with ck, x ∈R. We say that (Rudin [Rud64, Definition 3.38]): 5 •The series converges at x∈Rif limn→∞ Pn k=0 ckxkexists and is finite. •The radius of convergence R∈[0,+∞]is the supremum of the |x|for which the series converges. To state the Series Comparison Test, of enormous utility in Theorem 1: Theorem 1 (Comparison Test for Series).Let {ak}k∈N,{bk}k∈N⊂R+ 0be sequences of non-negative real numbers such that: 0≤bk≤ak∀k∈N. If the series P∞ k=1 akconverges, then P∞ k=1 bkalso converges. Proof. See p. 60 in Rudin [Rud64, Theorem 3.25]. Similarly, the concept of Asymptotic Upper Bound for a real function is defined through Definition 3: Definition 3 (Onotation (Asymptotic Upper Bound)).Let f, g:N→R(or f, g :R→R). We say that: f(x) = O(g(x)) as x→ ∞, if there exist constants x0>0and M > 0such that: |f(x)| ≤ M|g(x)|for all x≥x0. And its fundamental properties as follows in Proposition 3: Proposition 3 (Fundamental Properties of the ONotation).Let E⊆Rand let f1, f2, g1, g2:E→Rbe functions. For k∈R\ {0}, the following hold: (T) Transitivity:f1=O(g1)∧g1=O(g2) =⇒f1=O(g2); (M) Multiplication by Function (set equality): f2· O(g1)=O(f2g1); (S) Sum:f1=O(g1)∧f2=O(g2) =⇒f1+f2=O(max(|g1|,|g2|)); (H) Homogeneity:f1=O(g1) =⇒kf1=O(g1). Given all these preliminary concepts, this small section is finished, and we proceed to the next one. 2.1.2 On Pk∈N(Hk−γ)xk k! Two fundamental results for this work are proposed, which are stated below: Proposition 4. Let x∈R+, the following holds: X k∈N (Hk−γ)xk k!=exlog x+γ+O1 xas x→ ∞, where: 6 •N={1,2,3, . . .}denotes the set of natural numbers; •Hk=Pk n=1 1 nis the k-th harmonic number (H0:= 0); •γis the Euler-Mascheroni constant. Proof. See in Espinosa [Esp25a] (A Beautiful Equivalence of the Riemann Hypothesis). Proposition 5. For x∈R+, with x= 0, the following triple inequality holds: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!< exlog x+ log 1 + 1 x+γ. Proof. See in Espinosa [Esp25a] (A Beautiful Equivalence of the Riemann Hypothesis). We now have the series Pk∈N(Hk−γ)xk k!, with two fine bounds, an upper and a lower one respectively, and an asymptotic expression. Thus, everything required for it is already available, and therefore we proceed to the next section. 2.1.3 On the Espinosa-Riemann Equivalence Theorem In this Section 2.1.3, the result mentioned in the Introduction of this work is presented, which allows constructing equivalences of the Riemann Hypothesis! It is stated below: Theorem 2 (Espinosa-Riemann Equivalence Theorem).The Riemann Hypothesis is equivalent to: O(Hn) = X k∈N Hk n k!T(k)> σ(n) for all n∈N, where Hn=Pn i=1 1 iis the n-th harmonic number, σ(n) = Pd|nd is the sum-of-divisors function of n, and the functions O:R+→Rand T: R+→Rsatisfy the following conditions: 1. Asymptotic behavior of O(x): The function satisfies the asymptotic behavior: O(x):=X k∈N xk k!T(k)=exlog x+Oex xas x→ ∞. 2. Condition on the internal function for T(k): For all k∈Nthe inequality holds: T(k)> Hk−γ. 3. Validation condition for small n: The inequality O(Hn)> σ(n)holds for all natural numbers nin the range 1≤n≤60. 7 Proof. See in Espinosa [Esp25b] (On a Theorem of Equivalences of the Riemann Hypothesis). As can be seen in the definition of the function O(k), the function T(k)completely determines it and thus determines the equivalence in the same way. This means that if one wishes to construct an equivalence, all effort must be focused on adequately defining the function T(k), in order to meet the 3 requirements of the Theorem 2 and be valid. Thus, each T(k)induces an equivalence and vice versa, an equivalence can be obtained for each one, therefore they are defined in a bijective manner. Therefore, from now on, all effort will be focused on adequately constructing this function T(k). 2.1.4 On the Riemann zeta function ζand the Euler-Mascheroni constant γ It is now considered necessary to introduce the Riemann zeta function ζ. Unlike classical approaches, where it is presented to state the classical version of the Riemann Hypothesis, it is defined here to discretize the function T(k). A completely different approach, clearly. Definition 4 (Riemann zeta function ζ).For all s∈Cwith ℜ(s)>1, the Riemann zeta function ζis defined by the series: ζ(s):= ∞ X n=1 1 ns. This function admits analytic continuation to all C\ {1}(with a simple pole at s= 1) and satisfies: 1. Functional equation (symmetry): ζ(s) = 2sπs−1sin πs 2Γ(1 −s)ζ(1 −s); 2. Euler product (relation to primes): ζ(s) = Y pprime 1 1−p−sfor ℜ(s)>1; 3. Integral representation: ζ(s) = 1 Γ(s)Z∞ 0 xs−1 ex−1dx for ℜ(s)>1; 4. Special values: ζ(2) = π2 6, ζ(4) = π4 90, ζ(0) = −1 2, ζ(−1)=−1 12. 8 We proceed to relate the function ζto γ, which is vital for this work. As can be noted, the constant γhas a direct, if not deeply close, relationship with the Riemann zeta function ζ, as observed below in the following proposition: Proposition 6 (Relationship between γand the function ζ).The Euler-Masche roni constant γcan be expressed by the following series involving values of the Riemann zeta function ζat positive integers: γ= ∞ X n=2 (−1)nζ(n) n, where ζ(k)is the Riemann zeta function, defined for all k∈Cwith ℜ(k)>1, as ζ(k) = P∞ n=1 1 nkevaluated at n≥2. 2.1.5 On some useful inequalities In this part, certain inequalities are presented that will be extremely useful later on. Proposition 7. For r≥4and for x∈[2, r −2], the following holds: 1 x(r−x)x≤1 2(r−2)2. Proof. Since both sides of the inequality are positive, the proof is equivalent to demonstrating that the function f(x)=x(r−x)xsatisfies f(x)≥2(r−2)2in the interval [2, r −2]. We consider the auxiliary function k(x) = log f(x) = log x+xlog(r−x). We seek the minimum value of k(x)in the interval. First, we calculate the first and second derivatives of k(x): k′(x) = 1 x+ log(r−x) + x−1 r−x =1 x+ log(r−x)−x r−x k′′(x)=−1 x2−1 r−x−(1)(r−x)−x(−1) (r−x)2 =−1 x2−1 r−x−r (r−x)2. For r≥4and x∈[2, r −2], we have x>0and r−x≥2>0. Consequently, k′′(x)is always negative for x∈(2, r −2). Since k′′(x)<0, the function k(x) is concave in [2, r −2]. Therefore, the minimum value of k(x)must be attained at one of the endpoints of the interval: min{k(2), k(r−2)}. We evaluate k(x) at the endpoints: k(2) = log 2 + 2 log(r−2) = log(2(r−2)2) 9 Proposition 13 (Refinement of the Bounding of the Sum η(k)).Let η(k)be the sum defined in Proposition 12. For k≥1,η(k)can be bounded as follows: Hk−γ−5 2(k+ 2) < η(k)< Hk−γ+3 k+ 1. Proof. The equivalence of η(k)and its initial bounds are established in Proposition 12. We start from these initial bounds for η(k): log(k+1)− k X i=1 1 (k+2−i)ik+2−i≤η(k)≤log(k+ 1) + k X i=1 1 (k+2−i)ik+2−i. To introduce the Euler-Mascheroni constant γ, we use Proposition 1 applied to k+ 1: log(k+ 1) = Hk−γ+Z∞ k+1 1 ⌊t⌋−1 tdt. Substituting this expression for log(k+ 1) into the bounds of η(k), we obtain: Hk−γ+Z∞ k+1 1 ⌊t⌋−1 tdt − k X i=1 1 (k+2−i)ik+2−i≤η(k) ≤Hk−γ+Z∞ k+1 1 ⌊t⌋−1 tdt + k X i=1 1 (k+2−i)ik+2−i. To further refine this bounding, we define E(k) = R∞ k+1 1 ⌊t⌋−1 tdt and R(k) = Pk i=1 1 (k+2−i)ik+2−i. From Proposition 2 (applied with k+ 1 instead of k), we have the bounds for E(k): 1 (k+ 1)(k+ 2) < E(k)<1 k+ 1. For R(k), we observe that R(k) = 1 k+1 +Pk i=2 1 (k+2−i)ik+2−i. The remaining sum Pk i=2 1 (k+2−i)ik+2−ican be bounded using Proposition 7. Setting X=k+2−i and r=k+ 2, for X∈[2, k], each term 1 X(k+2−X)Xsatisfies 1 X(k+2−X)X≤ 1 2((k+2)−2)2=1 2k2. Thus, Pk i=2 1 (k+2−i)ik+2−i≤(k−1) 1 2k2=k−1 2k2. Therefore, we have the following bounds for R(k): 1 k+ 1 < R(k)≤1 k+ 1 +k−1 2k2. Now, we combine these bounds for the error terms E(k)and R(k): 1. For the upper limit of the error term (E(k) + R(k)): E(k)+R(k)<1 k+ 1 +1 k+ 1 +k−1 2k2=2 k+ 1 +k−1 2k2. 16 For k≥1,k−1 2k2≤k 2k2=1 2k: E(k) + R(k)<2 k+ 1 +1 2k=4k+ (k+ 1) 2k(k+ 1) =5k+ 1 2k(k+ 1). Since 5k+ 1 ≤6kfor k≥1: E(k) + R(k)<6k 2k(k+ 1) =3 k+ 1. 2. For the lower limit of the error term (E(k)−R(k)): E(k)−R(k)>1 (k+ 1)(k+ 2) −1 k+ 1 +k−1 2k2. E(k)−R(k)>1−(k+ 2) (k+ 1)(k+ 2)−k−1 2k2=−k−1 (k+ 1)(k+ 2)−k−1 2k2=−1 k+ 2−k−1 2k2. For k≥1,k−1 2k2≤k 2k2=1 2k: E(k)−R(k)>−1 k+ 2 −1 2k=−2k−(k+ 2) 2k(k+ 2) =−3k−2 2k(k+ 2). Since −3k−2≥ −5kfor k≥1: E(k)−R(k)>−5k 2k(k+ 2) =−5 2(k+ 2). Therefore, the final refined bounding for η(k)is: Hk−γ−5 2(k+ 2) < η(k)< Hk−γ+3 k+ 1. This demonstrates that η(k)is bounded around Hk−γwith an error term of order O(1 k), as stated in Proposition 13. Proposition 14 (Upper Bound for Hk−γ).For every integer k≥2, the EulerMascheroni constant γand the k-th harmonic number Hk=Pk j=1 1 jsatisfy: Hk−γ < η(k):= k X m=1 (−1)m+1 mH(m) k−m+1, where H(m) n=Pn j=1 1 jmdenotes the n-th generalized harmonic number of order m. Proof. To prove Proposition 14, we use the identity for the Euler-Mascheroni constant γthat is derived from Proposition 6 by expanding the Riemann zeta function ζ: γ= ∞ X m=2 ∞ X j=1 (−1)m mjm. 17 For the purpose of this proof, let us define an auxiliary sum η′(k)as: η′(k):= k X m=2 (−1)m mH(m) k−m+1. Expanding the generalized harmonic number H(m) k−m+1, we can write η′(k)as: η′(k) = k X m=2 k−m+1 X j=1 (−1)m mjm. The difference γ−η′(k)can be expressed as the sum of the terms (−1)m mjmfor all pairs (m, j)such that m≥2,j≥1, and m+j > k + 1. We can reindex this sum by r:=m+j: γ−η′(k) = ∞ X r=k+2 r−1 X m=2 (−1)m m(r−m)m. Let ar:=Pr−1 m=2 (−1)m m(r−m)m, so that: γ−η′(k) = ∞ X r=k+2 ar. The sum for aris ar=(−1)2 2(r−2)2+(−1)3 3(r−3)3+· · ·+(−1)r−1 (r−1)(1)r−1. Considering the last term (for m=r−1), which is (−1)r−1 r−1. If ris odd, r−1is even, so this term is 1 r−1. Every other term in the sum for 2≤m≤r−2is of the form (−1)m m(r−m)m. For m= 2, the term is 1 2(r−2)2. By Proposition 7 (with xin the proposition being mand rin the proposition being the rof ar), it holds that 1 m(r−m)m≤1 2(r−2)2. The total number of terms with 2≤m≤r−2is r−3. Thus, the sum of the absolute values of these r−3terms is at most (r−3) 1 2(r−2)2. Furthermore, as 1 r−1>1 2(r−2) . And we note that r−3 2(r−2)2=1 2(r−2) ·r−3 r−2<1 2(r−2) . Therefore, 1 r−1>1 2(r−2) >r−3 2(r−2)2. The sum aris an alternating sum with a dominant positive term 1 r−1(for m=r−1) and the remaining terms form an alternating sum whose absolute value is less than the value of the dominant term. Therefore, ar>0for odd r. For the proof that a2r+a2r+1 >0for all r, for r < 6, the inequality can be verified explicitly. We will assume r≥6. Shifting the index in the second sum by 1, we obtain: a2r+a2r+1 = 2r−1 X m=2 (−1)m m(2r−m)m+ 2r−1 X m=1 (−1)m+1 (m+ 1)(2r−m)m+1 =1 2(2r−1)2+ 2r−1 X m=2 (−1)m((2r−m)(m+ 1) −m) m(m+ 1)(2r−m)m+1 . 18 Now, we extract some terms from the sum for m= 2,3and the term for m= 2r−1. Rearranging and extracting the terms: a2r+a2r+1 =1 2(2r−1)2+3(2r−2) −2 6(2r−2)3−4(2r−3) −3 12(2r−3)4−1 (2r−1)(2r) + 2r−2 X m=4 (−1)m((2r−m)(m+ 1) −m) m(m+ 1)(2r−m)m+1 . Consider the remaining sum, P2r−2 m=4 (−1)m((2r−m)(m+1)−m) m(m+1)(2r−m)m+1 . Each term in the sum is of the form (−1)m((2r−m)(m+1)−m) m(m+1)(2r−m)m+1 . The first term in the sum, for m= 4, is positive (since m= 4 is even). Its value is (2r−4)(4+1)−4 4(4+1)(2r−4)4+1 =10r−24 20(2r−4)5. The Proposition 9 indicates that always ((2r−x)(x+1)−x) x(x+1)(2r−x)x+1 ≤10r−24 20(2r−4)5for x≥4. The total number of terms with 4≤m≤2r−2is 2r−5. Thus, the sum of these 2r−5terms is at most (2r−5) 10r−24 20(2r−4)5and at least −(2r−5) 10r−24 20(2r−4)5. Following this argument we obtain the provided inequality: a2r+a2r+1 ≥1 2(2r−1)2+3(2r−2) −2 6(2r−2)3−4(2r−3) −3 12(2r−3)4−1 (2r−1)(2r) −(2r−5)(5r−12) 10(2r−4)5. According to Proposition 8, this expression is strictly positive for all r≥6. Since we have established that the expression is a lower bound for a2r+a2r+1 and that it is strictly positive for r≥6, we conclude that a2r+a2r+1 >0for r≥6. As it was already mentioned that it is verified explicitly for r < 6, it holds for all r. We have demonstrated that ar>0for odd rand that a2r+a2r+1 >0 for all r. This implies that the sum P∞ r=k+2 aris always positive. Then, from γ−η′(k) = P∞ r=k+2 ar, it follows that γ−η′(k)>0. Therefore, γ > η′(k). Now, to arrive at the desired inequality Hk−γ < η(k), recall the definition of η(k): η(k) = k X m=1 (−1)m+1 mH(m) k−m+1. We can relate η(k)to η′(k)as follows: η(k) = (−1)1+1 1H(1) k−1+1 + k X m=2 (−1)m+1 mH(m) k−m+1. Since H(1) k=Hk, we have: η(k)=Hk+ k X m=2 (−1)m+1 mH(m) k−m+1. 19 Recalling that η′(k) = Pk m=2 (−1)m mH(m) k−m+1, we can see the relationship in signs. If we multiply η′(k)by −1: −η′(k)=− k X m=2 (−1)m mH(m) k−m+1 = k X m=2 (−1)m+1 mH(m) k−m+1. Thus, we have the relationship: η(k)=Hk−η′(k). From the inequality γ > η′(k), it follows that −γ < −η′(k). Adding Hkto both sides of this inequality: Hk−γ < Hk−η′(k). Finally, substituting Hk−η′(k)with η(k), we obtain: Hk−γ < η(k). This completes the demonstration of Proposition 14. 2.1.7 On Ξ(x):=Pk∈N xk k!η(k) This Section presents the function O(x) = Ξ(x)defined with T(k) = η(k)for the construction of the first proposed equivalence of the Riemann Hypothesis in this work, along with all the vital results for this work that derive from it, which are stated consecutively, starting next: Definition 5 (ΞFunction).Let x∈R+, we define: Ξ(x):=X k∈N xk k!η(k) where: •η(k):=Pk m=1 (−1)m+1 mH(m) k−m+1; •H(m) n=Pn j=1 1 jmdenotes the n-th generalized harmonic number of order m. Theorem 3 (Monotone Convergence Theorem for Series).A series P∞ k=1 ak with non-negative terms, i.e., ak≥0for all k∈N, converges if and only if the sequence of its partial sums SN=PN k=1 akis bounded above. That is, there exists M∈Rsuch that SN≤Mfor all N∈N. Proposition 15. The series Ξ(x)converges for all x∈R+. Proof. Fix an arbitrary x>0. By Definition 5 we have the function Ξ(x): Ξ(x):=X k∈N xk k!η(k). 20 To prove its convergence, we will apply the Theorem 1 and the Theorem 3 several times. By Proposition 13, we know that for k≥1: η(k)< Hk−γ+3 k+ 1. Multiplying this inequality by xk k!(which is >0for x>0), we obtain: xk k!η(k)<xk k!(Hk−γ) + 3xk k!(k+ 1). We define the function akto bound the terms of Ξ(x)from above: ak:=xk k!(Hk−γ) + 3xk k!(k+ 1). It is verified that xk k!η(k)< akfor all k≥1. Furthermore, by Proposition 14, Hk−γ < η(k). Since it is a known fact that Hk−γ > 0for k≥1, it follows that η(k)>0for all k≥1. For the lower side, by Proposition 14, we know that Hk−γ < η(k). Moreover, it is a known fact that Hk−γ > 0for all k∈N(for k≥1). Therefore, for k≥2,η(k)> Hk−γ > 0. For k= 1,η(1) = H(1) 1= 1, and H1−γ= 1 −γ≈0.42 >0. Thus, η(k)>0for all k≥1. Multiplying by xk k!(which is >0), we have: 0<xk k!η(k). Combining the inequalities, we obtain the fundamental relationship for applying the Comparison Test: 0<xk k!η(k)< ak. Now, we analyze the convergence of the series Pk∈Nak. This series decomposes into two parts: X k∈N ak=X k∈N xk k!(Hk−γ) + X k∈N 3xk k!(k+ 1). Consider the first sum, Pk∈N xk k!(Hk−γ). By Proposition 5, this series is bounded above by the finite expression exlog x+ log 1 + 1 x+γfor x∈R+. Since the terms of the series xk k!(Hk−γ)are positive for k≥1(because Hk−γ > 0), and their partial sums are bounded above by a finite quantity, it follows that this series converges by the Theorem 3. For the second sum, Pk∈N 3xk k!(k+1) , we can establish an upper bound. For k≥1, we know that 1 k+1 ≤1, which implies: 3xk k!(k+ 1) ≤3xk k!. 21 The series Pk∈N 3xk k!= 3 Pk∈N xk k!= 3ex. This is a known series that converges for all x∈R. Since 0<3xk k!(k+1) ≤3xk k!and Pk∈N 3xk k!converges, by the Theorem 1 (with ak=3xk k!(k+1) and bk=3xk k!), the series Pk∈N 3xk k!(k+1) also converges. Since both series that compose Pk∈Nak(i.e., Pk∈N xk k!(Hk−γ)and Pk∈N 3xk k!(k+1) ) converge, their sum Pk∈Nakalso converges. Finally, having 0< xk k!η(k)< akand knowing that Pk∈Nakconverges, we apply the Theorem 1 once again (with akin the theorem being our akand bkbeing our xk k!η(k)). We conclude that Pk∈N xk k!η(k) = Ξ(x)converges. Since the conclusion holds for all x>0, by the arbitrariness of x,Ξ(x)converges for all x∈R+. Proposition 16. For x∈R+, it holds that Ξ(x)> exlog x. Proof. Fix an arbitrary x>0. Recall the definition of the function Ξ(x)from Definition 5: Ξ(x):=X k∈N xk k!η(k). By Proposition 14, we know that for every integer k≥2: Hk−γ < η(k). For the case k= 1,H1−γ= 1 −γ≈1−0.5772 = 0.4228. And η(1) = P1 m=1 (−1)m+1 mH(m) 1−m+1 =(−1)1+1 1H(1) 1= 1 ·1 = 1. Since 0.4228 <1, the inequality Hk−γ < η(k)also holds for k= 1. Therefore, we can state that the inequality Hk−γ < η(k)is valid for all k≥1. Multiplying both sides of the inequality Hk−γ < η(k)by xk k!(which is a positive term, since x>0), we obtain: (Hk−γ)xk k!< η(k)xk k!. Summing over all values of k∈N(from k= 1 to ∞), we establish the following relationship for the series: X k∈N (Hk−γ)xk k!<X k∈N η(k)xk k!. By Definition 5, the right side of this inequality is precisely Ξ(x): X k∈N (Hk−γ)xk k!<Ξ(x). 22 Now, consider the left part of this inequality. By Proposition 5, we know that for any x∈R+: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!. Combining the two inequalities obtained, we have the following chain: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!<Ξ(x). From this chain of inequalities, we can directly infer that: exlog x+1 2log 1 + 2 x+γ < Ξ(x). Since x∈R+, the terms 1 2log 1 + 2 xand γare both positive. Specifically, log 1 + 2 x>0because 1 + 2 x>1, and the Euler-Mascheroni constant γ≈ 0.57721 is positive. Therefore, the sum of these two terms, 1 2log 1 + 2 x+γ, is strictly greater than zero. This implies that: exlog x < exlog x+ 1 2log 1 + 2 x+γ!. Finally, by combining this last inequality with the one we established previously: exlog x < exlog x+1 2log 1 + 2 x+γ!<Ξ(x), we conclude that: exlog x < Ξ(x). This proof is valid for all x∈R+. Proposition 17. For all x∈R+, it holds: Ξ(x) = exlog x+Oex xas x→ ∞. Proof. Fix an arbitrary x > 0and consider the behavior of the function Ξ(x) as x→ ∞. Recall the definition of the function Ξ(x)from Definition 5: Ξ(x):=X k∈N xk k!η(k). To analyze the asymptotic behavior, we start with Proposition 13, which establishes the following bound for η(k)for k≥1: Hk−γ−5 2(k+ 2) < η(k)< Hk−γ+3 k+ 1. 23 Rearranging these inequalities for the difference η(k)−(Hk−γ): −5 2(k+ 2) < η(k)−(Hk−γ)<3 k+ 1. This implies that the absolute value of the difference is bounded: η(k)−(Hk−γ)<max 5 2(k+ 2),3 k+ 1. For k≥1, the term 3 k+1 is always greater than or equal to 5 2(k+2) . For example, for k= 1,3 2vs 5 6. As k→ ∞,3 k+1 ∼3 kand 5 2(k+2) ∼5 2k, and since 3>5/2, the inequality holds. Therefore, for k≥1: η(k)−(Hk−γ)<3 k+ 1. Now, we multiply this inequality by xk k!, which is positive for x > 0:  xk k!η(k)−xk k!(Hk−γ) <3xk k!(k+ 1). We sum over all values of k∈N(from k= 1 to ∞). By the triangle inequality (the sum of the absolute values is greater than or equal to the absolute value of the sum), we have: X k∈N xk k!η(k)−xk k!(Hk−γ)! ≤X k∈N xk k!η(k)−xk k!(Hk−γ) <X k∈N 3xk k!(k+ 1). The left side of this chain is Ξ(x)−Pk∈N(Hk−γ)xk k!. Let’s focus on the series on the right side: X k∈N 3xk k!(k+ 1). We observe that k!(k+ 1) = (k+ 1)!. Thus, the series becomes: 3 ∞ X k=1 xk (k+ 1)!. To simplify this sum, we perform an index change. Let j=k+ 1. When k= 1, j= 2. Then, k=j−1: 3 ∞ X j=2 xj−1 j!=3 x ∞ X j=2 xj j!. 24 We know that the Taylor series for exis ex=P∞ j=0 xj j!=x0 0! +x1 1! +P∞ j=2 xj j!= 1 + x+P∞ j=2 xj j!. Therefore, P∞ j=2 xj j!=ex−1−x. Substituting this into the previous expression: 3 ∞ X j=2 xj−1 j!=3 x(ex−1−x) = 3ex x−3 x−3. Thus, we have established the following bound:  Ξ(x)−X k∈N (Hk−γ)xk k! <3ex x−3 x−3. For sufficiently large x, the expression 3ex x−3 x−3is positive and is bounded above by 3ex x(since 3 x+ 3 >0). By Definition 3, if we consider f(x) = Ξ(x)−Pk∈N(Hk−γ)xk k!and g(x) = ex x, we can observe that for sufficiently large x(there exists x0such that for all x≥x0,3ex x−3 x−3<3ex x), the inequality satisfies the condition for the Onotation:  Ξ(x)−X k∈N (Hk−γ)xk k! ≤M ex x for a constant M(for example, M= 3). Therefore, we conclude that: Ξ(x)−X k∈N (Hk−γ)xk k!=Oex xas x→ ∞. Now, we consider Proposition 4, which gives us the asymptotic behavior of the series Pk∈N(Hk−γ)xk k!: X k∈N (Hk−γ)xk k!=exlog x+γ+O1 xas x→ ∞. Let’s analyze the terms O1 xand γ. We know that limx→∞ 1/x ex/x = limx→∞ 1 ex= 0. This means that 1 x=oex x, and by Definition 3, if f(x)=o(g(x)), then f(x)=O(g(x)). Thus, O1 xis of a lower order and, therefore, can be absorbed by Oex x. The constant γis a constant term, and limx→∞ γ ex/x = limx→∞ γx ex= 0. This implies that γ=oex x, and therefore, γ=Oex x. Applying the Sum (S) property of Proposition 3, if we have f1(x)=γ(which is Oex x) and f2(x)=O1 x(which is also Oex x), then their sum is: γ+O1 x=O max |γ|, 1 x! . 25 Figure 3: Bound of Espinosa’s Inequality from Theorem 4 for the function σ(n), for values of n∈Nwith n≤10000. Figure 4: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 4 for the function σ(n), for values of k∈Nwith k≤100. 32 Figure 5: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 4 for the function σ(n), for values of k∈Nwith k≤1000. 2.1.9 On another important relationship between the Riemann zeta function ζand the Euler-Mascheroni constant γ The necessary results for the construction of the second equivalence of the Riemann Hypothesis, presented in this work, are shown here. Proposition 18 (Version of the Euler-Mascheroni constant γin terms of the positive Riemann zeta function ζ).The Euler-Mascheroni constant γcan be expressed by the following series involving the values of the Riemann zeta function ζat positive integers: γ= 1 − ∞ X k=2 ζ(k)−1 k where ζ(k)is the Riemann zeta function, defined for all k∈Cwith ℜ(k)>1, as ζ(k) = P∞ n=1 1 nk. Proposition 19 (Dirichlet eta function η(Alternating Riemann zeta function ζSeries)).The Dirichlet eta function η(s)is defined for ℜ(s)>0as the alternating series: η(s) = ∞ X n=1 (−1)n−1 ns. For ℜ(s)>1, the Dirichlet eta function is related to the Riemann zeta function ζ(s)by the following identity: η(s) = (1 −21−s)ζ(s) 33 or, equivalently, ζ(s) = 1 1−21−sη(s). 2.1.10 On other useful inequalities We continue to state another set of essential inequalities, each result in an uninterrupted manner, similarly: Proposition 20 (Taylor’s Remainder Theorem (Lagrange Form)).Let f(x)be a function that has (N+ 1) continuous derivatives on an interval Icontaining a. The value of the function f(x)can be approximated by its degree NTaylor polynomial around a, denoted PN(x): PN(x) = N X k=0 f(k)(a) k!(x−a)k. The truncation error, or Lagrange remainder, RN(x)=f(x)−PN(x), is given by: RN(x) = f(N+1)(c) (N+ 1)! (x−a)N+1 for some value cthat lies strictly between aand x. Proposition 21 (Bound for the Partial Sum of the Natural Logarithm Taylor Series (Positive Terms Form)).For any natural number n≥1, the function log(1+ 1 n)can be expressed by the following Taylor series around x= 0, evaluated at x=1 n+1 : S= log 1 + 1 n= ∞ X m=1 1 m(n+ 1)m. If Sj(n) = Pj m=1 1 m(n+1)mdenotes the j-th partial sum of this series, then Sj(n) is bounded by: log 1 + 1 n−1 (j+ 1)nj+1 ≤Sj(n)≤log 1 + 1 n. Proof. Consider the Taylor series for f(x) = −log(1 −x)evaluated at x=1 n+1 : S=−log 1−1 n+ 1=−log n n+ 1= log n+ 1 n= log 1 + 1 n. The Maclaurin series expansion for f(x)=−log(1 −x)is: −log(1 −x) = ∞ X m=1 xm mfor |x|<1. 34 Evaluating at x=1 n+1 , we obtain the series of positive terms: S= ∞ X m=1 1 m(n+ 1)m. To bound the truncation error, we will use Taylor’s Remainder Theorem (Proposition 20). Let Sj(n)be the j-th partial sum of the series. The truncation error, Rj(n), is defined as Rj(n)=S−Sj(n). The formula for the Taylor remainder for f(x)around a= 0 is: Rj(x) = f(j+1)(c) (j+ 1)! xj+1 where cis some value between 0and x. First, we calculate the (j+ 1)-th derivative of f(x)=−log(1 −x): f′(x) = 1 1−x= (1 −x)−1 f′′(x) = (1 −x)−2 f′′′(x) = 2(1 −x)−3 . . . f(k)(x)=(k−1)!(1 −x)−k. Thus, the (j+ 1)-th derivative is: f(j+1)(x)=j!(1 −x)−(j+1). Now, we substitute this into the remainder formula, with x=1 n+1 : Rj1 n+ 1=j!(1 −c)−(j+1) (j+ 1)! 1 n+ 1j+1 . Simplifying: Rj1 n+ 1=1 j+ 1 1 (1 −c)j+1 1 (n+ 1)j+1 . Since 0< c < 1 n+1 (for n≥1), the term 1−cis in the interval 1−1 n+1 ,1= n n+1 ,1. To bound the error above, we need to maximize the term 1 (1−c)j+1 . Since the function g(c) = (1 −c)−(j+1) is increasing in cfor c < 1, its maximum value on the interval (0,1 n+1 )occurs when capproaches 1 n+1 . Thus, the maximum value of 1 (1−c)j+1 is 1 (1−1 n+1 )j+1 =1 (n n+1 )j+1 =n+1 nj+1. Substituting this upper limit into the expression for Rj: |Rj(n)| ≤ 1 j+ 1 n+ 1 nj+1 1 (n+ 1)j+1 35 |Rj(n)| ≤ 1 j+ 1 (n+ 1)j+1 nj+1 1 (n+ 1)j+1 . Simplifying, we obtain the upper bound of the error: |Rj(n)| ≤ 1 (j+ 1)nj+1 . Since all terms of the series P∞ m=1 1 m(n+1)mare positive, the partial sum Sj(n) will always be less than the total sum S(unless jis infinite). This means that the error Rj(n)=S−Sj(n)is always positive. Therefore, we can write the inequality without absolute value and with the lower bound of zero: 0≤Rj(n)≤1 (j+ 1)nj+1 . Substituting Rj(n) = log 1 + 1 n−Sj(n): 0≤log 1 + 1 n−Sj(n)≤1 (j+ 1)nj+1 . From the first part of the inequality, 0≤log 1 + 1 n−Sj(n), we obtain: Sj(n)≤log 1 + 1 n. From the second part of the inequality, log 1 + 1 n−Sj(n)≤1 (j+1)nj+1 , we obtain: Sj(n)≥log 1 + 1 n−1 (j+ 1)nj+1 . Combining both inequalities, we obtain the desired bound for the partial sum: log 1 + 1 n−1 (j+ 1)nj+1 ≤Sj(n)≤log 1 + 1 n. This completes the demonstration of the partial sum bounds for the positive terms series. Proposition 22. For every integer k≥2, the expression: Ck= 1 − k X p=2 1 p(h(p) k+1) satisfies the inequality: γ < Ck<γ+1 k+ 1 where γis the Euler-Mascheroni constant and h(p) n=Pn j=2 1 jp. 36 Proof. We start with the formula of Proposition 18 for the Euler-Mascheroni constant γ: γ= 1 − ∞ X p=2 1 p(ζ(p)−1). Since ζ(p)−1 = P∞ j=2 1 jp, we have: γ= 1 − ∞ X p=2 1 p  ∞ X j=2 1 jp . We split the infinite summation over pinto two parts: up to k, and from k+ 1 onwards: γ=  1− k X p=2 1 p  ∞ X j=2 1 jp   −   ∞ X p=k+1 1 p  ∞ X j=2 1 jp   . Next, we expand the inner summation P∞ j=2 in the first bracket, using the partition Pk+1 j=2 +P∞ j=k+2: γ=  1− k X p=2 1 p  k+1 X j=2 1 jp+ ∞ X j=k+2 1 jp   − ∞ X p=k+1 ∞ X j=2 1 pjp. Distributing and grouping terms, we identify Ckand define the truncation error Ek: γ= 1− k X p=2 1 p k+1 X j=2 1 jp  | {z } Ck −  k X p=2 ∞ X j=k+2 1 pjp+ ∞ X p=k+1 ∞ X j=2 1 pjp  | {z } Ek . The exact relationship is established: Ck=γ+Ekwhere Ek>0. I. Proof of the Lower Bound (γ < Ck): Substituting the key relationship: γ < γ +Ek. Since Ekis a sum of strictly positive terms ( 1 pjpwith p≥2, j ≥2), the condition Ek>0holds strictly. Thus, the lower bound is proven. II. Proof of the Upper Bound (Ck<γ+1 k+1 ): The upper bound γ+Ek< γ +1 k+1 is equivalent to proving the inequality: Ek<1 k+1 . We rearrange Ek, setting n=k+ 1: Ek= n X j=2 ∞ X p=n 1 pjp | {z } A + ∞ X j=n+1 ∞ X p=2 1 pjp | {z } B . 37 1. Bounding the Second Sum (B): For fixed j≥n+ 1 ≥4, the inner sum satisfies P∞ p=2 1 pjp<P∞ p=2 1 2jp(since 1 p≤1 2, with strict inequality for p>2). Using the geometric series formula P∞ p=2 rp=r2 1−rwith r=1 j: ∞ X p=2 1 pjp<1 2·(1 j)2 1−1 j =1 2j(j−1). Substituting into B(telescoping series): B < ∞ X j=n+1 1 2j(j−1) =1 2 ∞ X j=n+1 1 j−1−1 j=1 2·1 n. 2. Bounding the First Sum (A): We bound the inner sum Pn j=2 1 jpstrictly by 1 2p+ (n−2) 1 3p. Substituting this into A: A < ∞ X p=n 1 p·1 2p+ (n−2) ∞ X p=n 1 p·1 3p. We apply the strict bound for the tail of the log series: P∞ p=nxp p<1 n·xn 1−x. 1. For x=1 2:P∞ p=n (1 2)p p<(1 2)n−1 n. 2. For x=1 3:P∞ p=n (1 3)p p<3 2n1 3n. Substituting these into the expression for A: A < (1 2)n−1 n+ (n−2) 3 2n1 3n . Combining the strict bounds for Aand B: Ek=A+B < 1 n"1 2n−1 +3 2(n−2) 1 3n +1 2#. To show Ek<1 n, we must prove that the term in the bracket is strictly less than 1, or f(n)<1 2, where: f(n) = 1 2n−1 +3 2(n−2) 1 3n . 1. Base Case (n= 3): f(3) = 1 4+1 18 =11 36 . Since 11 36 <1 2(or 18 36 ), the inequality holds. 2. Monotonicity (n≥3): The difference f(n+1)−f(n)simplifies to: f(n+1)−f(n)=−1 2n+5−2n 2·3n. Since n≥3,5−2nis negative, proving f(n+1)−f(n)<0. 38 f(n)is strictly decreasing, thus f(n)< f(3) <1 2for all n≥3. Substituting this result back into the bound for Ek: Ek<1 nf(n) + 1 2<1 n1 2+1 2=1 n=1 k+ 1. The inequality Ek<1 k+1 is proven strictly. III. Conclusion Substituting the result of the upper bound proof (Ek<1 k+1 ) into the key relationship Ck=γ+Ekwe have: γ < Ck<γ+1 k+ 1. 2.1.11 On τ(k)and its upper and lower bounds This Section 2.1.11 presents the function τ(k), which is simply the function T(k) from the second equivalence. As stated in Section 2.1.3, it entirely characterizes the second equivalence, analogous to how η(k)characterizes the first equivalence. Similar to the latter, this Section 2.1.11 for τ(k)is presented like Section 2.1.6 for η(k), stating a set of necessary results to bound and provide the general and asymptotic behavior of τ(k). They are stated below, consecutively and without interruption: Definition 6. Let h(p) nbe the n-th generalized harmonic number of order p starting from j= 2, defined as: h(p) n= n X j=2 1 jp. Then, the function τ(k)is defined for each natural number k≥1as the following sum: τ(k) = k X i=1 1 ih(i) k+1. Proposition 23 (Equivalence and Bound for the Sum τ(k)).Let τ(k)be the sum defined as: τ(k):= k X i=1 Sk+1−i(i)=Sk(1) + Sk−1(2) + ···+S1(k). Where Sj(n) = Pj m=1 1 m(n+1)mis the j-th partial sum of the Taylor series for log(1 + 1 n)in its positive terms form. This sum τ(k)is equivalent to the expression: τ(k) = k X m=1 1 m˜ H(m) k−m+1 39 where ˜ H(m) n=Pn j=1 1 (j+1)mare the modified generalized harmonic numbers, in which each term of the sum is 1 (j+1)minstead of 1 jm. Furthermore, τ(k)can be bounded by: log(k+1)− k X i=1 1 (k+2−i)ik+2−i≤τ(k)≤log(k+ 1). Proof. We start with the definition of τ(k)as a sum of partial sums: τ(k) = k X i=1 Sk+1−i(i). Now, we substitute the definition of Sj(n)into the sum, where j=k+1−i and n=i. It is crucial to remember that for this τ(k),Sj(n)is based on the positive terms series, where the denominators are (n+ 1)m: τ(k) = k X i=1   k+1−i X m=1 1 m(i+ 1)m . To demonstrate the equivalence with the desired form involving harmonic numbers, we interchange the order of the summations. The domain of the indices is 1≤i≤kand 1≤m≤k+ 1 −i. From the second inequality, i≤k+ 1 −m. Since i≥1, this implies 1≤k+1−m, or m≤k. Thus, mranges from 1to k, and for each m,iranges from 1to k+ 1 −m. τ(k) = k X m=1 k+1−m X i=1 1 m(i+ 1)m. We factor out the term that does not depend on i: τ(k) = k X m=1 1 m  k+1−m X i=1 1 (i+ 1)m . The internal sum Pk+1−m i=1 1 (i+1)mis a modified form of the generalized harmonic numbers. We define it as ˜ H(m) k+1−m. Therefore, we have demonstrated the equivalence: τ(k) = k X m=1 1 m˜ H(m) k+1−m. We use the inequalities for Sj(n)established in Proposition 21 (the positive terms version): log 1 + 1 n−1 (j+ 1)nj+1 ≤Sj(n)≤log 1 + 1 n. 40 For each term Sk+1−i(i)in the sum τ(k), where j=k+1−iand n=i, we substitute these values into the inequalities: log 1 + 1 i−1 ((k+1−i) + 1)i((k+1−i)+1) ≤Sk+1−i(i)≤log 1 + 1 i. Simplifying the denominators and exponents: log 1 + 1 i−1 (k+2−i)ik+2−i≤Sk+1−i(i)≤log 1 + 1 i. Now, we sum these inequalities term by term for ifrom 1to k: k X i=1 log 1 + 1 i−1 (k+2−i)ik+2−i!≤ k X i=1 Sk+1−i(i)≤ k X i=1 log 1 + 1 i. We separate the sums: k X i=1 log 1 + 1 i− k X i=1 1 (k+2−i)ik+2−i≤τ(k)≤ k X i=1 log 1 + 1 i. The sum of logarithms is a well-known telescoping sum: k X i=1 log 1 + 1 i= k X i=1 log i+ 1 i = (log(2) −log(1)) + ···+ (log(k+1)−log(k)) = log(k+ 1) −log(1) = log(k+ 1). Substituting this result into the inequality: log(k+1)− k X i=1 1 (k+2−i)ik+2−i≤τ(k)≤log(k+ 1). This is the bound for τ(k). Proposition 24 (Refinement of the Bound for the Sum τ(k)).Let τ(k)be the sum defined in Proposition 23. For k≥1,τ(k)can be bounded as follows: Hk−γ−5 2(k+ 2) < τ(k)< Hk−γ+1 k+ 1. Proof. The initial bounds for τ(k)are established in Proposition 23: log(k+1)− k X i=1 1 (k+2−i)ik+2−i≤τ(k)≤log(k+ 1). 41 Combining the two inequalities obtained, we have the following chain: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!< κ(x). From this chain of inequalities, we can directly infer that: exlog x+1 2log 1 + 2 x+γ < κ(x). Since x∈R+, the terms 1 2log 1 + 2 xand γare both positive. Specifically, log 1 + 2 x>0because 1 + 2 x>1, and the Euler-Mascheroni constant γ≈ 0.57721 is positive. Therefore, the sum of these two terms, 1 2log 1 + 2 x+γ, is strictly greater than zero. This implies that: exlog x < exlog x+ 1 2log 1 + 2 x+γ!. Finally, by combining this last inequality with the one we established previously: exlog x < exlog x+1 2log 1 + 2 x+γ!< κ(x), we conclude that: exlog x < κ(x). This proof is valid for all x∈R+. Proposition 28. For all x∈R+, it holds: κ(x) = exlog x+Oex xas x→ ∞. Proof. Let us fix an arbitrary x > 0and consider the behavior of the function κ(x)as x→ ∞. Recall the definition of the function κ(x)from Definition 7: κ(x):=X k∈N xk k!τ(k). To analyze the asymptotic behavior, we start with Proposition 25, which establishes the following bound for τ(k)for k≥1: Hk−γ < τ(k)< Hk−γ+1 k+ 1. Rearranging these inequalities for the difference τ(k)−(Hk−γ): 0< τ(k)−(Hk−γ)<1 k+ 1. 48 This implies that the absolute value of the difference is bounded: τ(k)−(Hk−γ)<1 k+ 1. Now, we multiply this inequality by xk k!, which is positive for x > 0:  xk k!τ(k)−xk k!(Hk−γ) <xk k!(k+ 1). We sum over all values of k∈N(from k= 1 to ∞). By the triangle inequality (the sum of the absolute value is greater than or equal to the absolute value of the sum), we have: X k∈N xk k!τ(k)−xk k!(Hk−γ)! ≤X k∈N xk k!τ(k)−xk k!(Hk−γ) <X k∈N xk k!(k+ 1). The left side of this chain is κ(x)−Pk∈N(Hk−γ)xk k!. Let’s focus on the series on the right side: X k∈N xk k!(k+ 1). We observe that k!(k+ 1) = (k+ 1)!. Thus, the series becomes: ∞ X k=1 xk (k+ 1)!. To simplify this sum, we perform an index change. Let j=k+ 1. When k= 1, j= 2. Then, k=j−1:∞ X j=2 xj−1 j!=1 x ∞ X j=2 xj j!. We know that the Taylor series for exis ex=P∞ j=0 xj j!=x0 0! +x1 1! +P∞ j=2 xj j!= 1 + x+P∞ j=2 xj j!. Therefore, P∞ j=2 xj j!=ex−1−x. Substituting this into the previous expression: ∞ X j=2 xj−1 j!=1 x(ex−1−x) = ex x−1 x−1. Thus, we have established the following bound:  κ(x)−X k∈N (Hk−γ)xk k! <ex x−1 x−1. 49 For sufficiently large x, the expression ex x−1 x−1is positive and bounded above by ex x(since 1 x+ 1 >0). By Definition 3, if we consider f(x) = κ(x)−Pk∈N(Hk−γ)xk k!and g(x) = ex x, we can observe that for sufficiently large x(there exists x0such that for all x≥x0,ex x−1 x−1<ex x), the inequality satisfies the condition for the Onotation:  κ(x)−X k∈N (Hk−γ)xk k! ≤M ex x for a constant M(for example, M= 1). Therefore, we conclude that: κ(x)−X k∈N (Hk−γ)xk k!=Oex xas x→ ∞. Now, we consider Proposition 4, which gives us the asymptotic behavior of the series Pk∈N(Hk−γ)xk k!: X k∈N (Hk−γ)xk k!=exlog x+γ+O1 xas x→ ∞. Let’s analyze the terms O1 xand γ. We know that limx→∞ 1/x ex/x = limx→∞ 1 ex= 0. This means that 1 x=oex x, and by Definition 3, if f(x)=o(g(x)), then f(x)=O(g(x)). Thus, O1 xis of a lower order and can therefore be absorbed by Oex x. The constant γis a constant term, and limx→∞ γ ex/x = limx→∞ γx ex= 0. This implies that γ=oex x, and therefore, γ=Oex x. Applying the Sum (S) property of Proposition 3, if we have f1(x)=γ(which is Oex x) and f2(x)=O1 x(which is also Oex x), then their sum is: γ+O1 x=O max |γ|, 1 x! . Since both are oex x(they tend to zero faster than ex x), when we sum them, the combined sum is still oex x, which implies that the sum is Oex x. Therefore, the previous expression can be rewritten as: X k∈N (Hk−γ)xk k!=exlog x+Oex xas x→ ∞. Finally, we combine the two expressions for κ(x): κ(x) =  X k∈N (Hk−γ)xk k! + κ(x)−X k∈N (Hk−γ)xk k! . 50 Substituting the asymptotic expressions we found: κ(x) = exlog x+Oex x!+Oex x. Applying the Sum (S) property of Proposition 3 again, since both Oex x terms have the same order of magnitude, their sum is also Oex x: κ(x) = exlog x+Oex x. The proof is valid as x→ ∞. 2.1.13 Espinosa’s Riemann Hypothesis Equivalence (2025) with T(k) = τ(k) Finally the second equivalence of the coveted Riemann Hypothesis is stated in the Theorem 5 as follows: Theorem 5 (Espinosa, 2025).The Riemann Hypothesis is equivalent to: X k∈N Hk n k!τ(k)> σ(n) for all n∈N, where: •τ(k) = Pk i=1 1 ih(i) k+1; •h(p) m=Pm j=2 1 jpdenotes the m-th generalized harmonic number of order p starting from j= 2; •Hn=Pn i=1 1 iis the n-th harmonic number; •σ(n) = Pd|ndis the sum-of-divisors function of n. Proof. By Theorem 2, taking the functions O(x) = κ(x)and T(x) = τ(k), and since the three required conditions for its application are met, respectively, the condition of Asymptotic behavior of O(x)by Proposition 28, the Condition on the inner function for T(k)by Proposition 27 and the Validation condition for small nis verified, since the inequality O(Hn)> σ(n)holds for all natural numbers nin the range 1≤n≤60. Then the desired result follows. An explicit version without the use of the function τ(k)is presented, truly beautiful! 51 Corollary 5 (Espinosa, 2025).The Riemann Hypothesis is equivalent to: H1 n 1! 1 1h(1) 2+H2 n 2! 1 1h(1) 3+1 2h(2) 3+H3 n 3! 1 1h(1) 4+1 2h(2) 4+1 3h(3) 4+···> σ(n) for all n∈N, where: •h(p) m=Pm j=2 1 jpdenotes the m-th generalized harmonic number of order p starting from j= 2; •Hn=Pn i=1 1 iis the n-th harmonic number; •σ(n) = Pd|ndis the sum-of-divisors function of n. Proof. The equivalence of the Riemann Hypothesis with a sum of the form Pk∈N Hk n k!ak> σ(n)is a result derived from Theorem 5. In that theorem, the coefficient akis defined as ak:=τ(k). The series presented in the statement of this theorem is merely the explicit form of the first terms of this summation, obtained by expanding the definition of τ(k)for k= 1,2,3: •For k= 1: a1=τ(1) = P1 i=1 1 ih(i) 1+1=1 1h(1) 2. •For k= 2: a2=τ(2) = P2 i=1 1 ih(i) 2+1=1 1h(1) 3+1 2h(2) 3. •For k= 3: a3=τ(3) = P3 i=1 1 ih(i) 3+1=1 1h(1) 4+1 2h(2) 4+1 3h(3) 4. The coefficients of the subsequent terms in the summation (···) are derived in the same manner, extending the definition of τ(k)for increasing values of k. Therefore, the explicit representation of the series is a direct consequence of the definition of the coefficients of the summation equivalent to the Riemann Hypothesis. Finally, a recursive form for the coefficient ak, seeing the equivalence expression as Pk∈N Hk n k!ak. Corollary 6 (Espinosa, 2025).The Riemann Hypothesis is equivalent to: X k∈N Hk n k!ak> σ(n) for all n∈N, where: •The coefficient akis defined recursively as: ak=ak−1+Pk−1 i=1 1 i(k+1)i+1 kh(k) k+1 for k > 1, with a1=1 2; 52 •h(p) m=Pm j=2 1 jpdenotes the m-th generalized harmonic number of order p starting from j= 2; •Hn=Pn i=1 1 iis the n-th harmonic number; •σ(n) = Pd|ndis the sum-of-divisors function of n. Proof. The equivalence of the Riemann Hypothesis with the presented inequality is established from the result that uses the function τ(k)from Theorem 5. To complete the proof of this theorem, it is necessary to verify that the coefficient ak, defined recursively in the statement, is indeed the function τ(k), which validates its use in the equivalence. Recall the explicit definition of the function τ(k): τ(k):= k X i=1 1 ih(i) k+1 where h(p) m=Pm j=2 1 jpdenotes the generalized harmonic number starting from j= 2. Now, we will demonstrate that τ(k)satisfies the recursive relation given for akin the theorem statement. Consider the difference τ(k)−τ(k−1) for k > 1: τ(k)−τ(k−1) =   k X i=1 1 ih(i) k+1 −  k−1 X i=1 1 ih(i) k . We separate the last term of the first sum: τ(k)−τ(k−1) = k−1 X i=1 1 ih(i) k+1 +1 kh(k) k+1 − k−1 X i=1 1 ih(i) k. We regroup the sums: τ(k)−τ(k−1) = k−1 X i=1 1 i(h(i) k+1 −h(i) k) + 1 kh(k) k+1. We use the recursive property of the generalized harmonic numbers starting from j= 2:h(p) n−h(p) n−1=1 np. Applying this to h(i) k+1 −h(i) k, we have: h(i) k+1 −h(i) k=1 (k+ 1)i. Substituting this expression into the difference: τ(k)−τ(k−1) = k−1 X i=1 1 i 1 (k+ 1)i+1 kh(k) k+1. 53 Therefore, the recursive relation for τ(k)is: τ(k) = τ(k−1) + k−1 X i=1 1 i(k+ 1)i+1 kh(k) k+1 for k > 1. The initial condition for τ(k)is obtained directly from its explicit definition for k= 1: τ(1) = 1 X i=1 1 ih(i) 1+1=1 1h(1) 2=1 2. Since the coefficient akin the theorem statement is defined with this same recursive relation and initial condition, it is concluded that ak≡τ(k). With this, the equivalence of the Riemann Hypothesis is maintained, and a recursive form for the calculation of the coefficients is provided. Figure 6: Bound of Espinosa’s Inequality from Theorem 5 for the function σ(n), for values of n∈Nwith n≤100. 54 Figure 7: Bound of Espinosa’s Inequality from Theorem 5 for the function σ(n), for values of n∈Nwith n≤1000. Figure 8: Bound of Espinosa’s Inequality from Theorem 5 for the function σ(n), for values of n∈Nwith n≤10000. 55 Figure 9: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 5 for the function σ(n), for values of k∈Nwith k≤100. Figure 10: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 5 for the function σ(n), for values of k∈Nwith k≤1000. 56 References [Apo76] T. M. Apostol. Introduction to Analytic Number Theory. 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